Question

Is anyone here the master of work and energy and can help a freshman out? Regarding spins

Answer 1

There's a 2kg mass moving on a surface which in the end there's a spin. The distance between A-B is 5 meters and ONLY between A-B there's friction (meaning under the spin there's no friction) What should be the initial velocity of the mass so it will come back to point A? the coefficient of friction is 0.2.

Answer 2

I shall cherish you all whoever is capable of solving it!!!!

Answer 3

I assume the spring is fixed in place and the 2 Kg mass compresses it and the spring stopping it will reverse its motion and the mass leaves the spring with the same speed as it engaged it. The initial KE energy must equal the work done in overcoming friction in both directions.

Answer 4

I think that the initial kinetic energy of your mass has to be greater than the work by the friction force acting in both direction, namely:
\[\frac{ 1 }{ 2 }mv ^{2}>2(\mu*m*g*d)\]
where \[\mu \]
is the coefficient of sliding friction

Answer 5

of course I suppose that the collision between the mass and the spring is elastic

Answer 6

Thank you! but what the the answer then?

Answer 7

Approximately 6.26 m/s

Answer 8

you're brilliant!!! Can you please show me the solution?

Answer 9

So I will be able to do it myself for next time

Answer 10

The problem with assuming that the kinetic energy must be greater than the work done to over come friction in both directions is that the mass would land at a position beyond A.

Set the equation for kinetic energy equal to 2 * the equation for the friction force:
\[1/2 *mv^2 = 2(\mu m g d)\]

Answer 11

plug in everything:
1/2*(2kg)v^2 = 2[(0.2)(2kg)(9.81m/s^2)(5m)]

Answer 12

then solve for v

Answer 13

You are truly my hero!!!!! Thank you so much!!!!!!!!!!!!!!!

Answer 14

no problem

Answer 15

Can I ask you something short?

Answer 16

sure

Answer 17

Based on the K of the spin is 52.3N, what would be the Delta X of the spin when the box hits it? Which equation should I use?

Answer 18

when you say spin, are you referring to a spring?

Answer 19

0.skx^2=W?

Answer 20

spring yes sorry

Answer 21

so if the spring constant, k, is 52.3 N/m, are you asking what the force is when the box hits it?

Answer 22

what would be the constriction of the spring when the box hits it

Answer 23

*hits the spring

Answer 24

Use hooke's law and solve for X

Answer 25

I tried to do it and got confused...

Answer 26

KX^2*0.5=W

Answer 27

Am I correct? And to work it will be coefficient of friction*mg*5*cos180?

Answer 28

are you talking about the elastic potential energy equation?

Answer 29

I thought that would fit the best here, am I completely wrong?

Answer 30

set that equal to kx and solve for x

Answer 31

0.5KX^2-0.5MV^2=coefficient of friction*mg*cos180*5 meters?

Answer 32

Is this the calculation I should do?

Answer 33

BTW thank you so much for everything

Answer 34

Am I right this time?

Answer 35

?

Answer 36

I think I've got it. Thanks allot!!!!! You've truly made my day brighter!

Answer 37

But can you just approve my equation?

Answer 38

?

Answer 39

Your equation is correct and can be written as \[\frac{ 1 }{ 2 }kx ^{2} =\frac{ 1 }{ 2 }mv ^{2}-5\mu mg\]where v is the initial velocity of the box, x is the net compression of the spring.

The equation states that the energy stored in the spring is equal to the net energy available just before the spring and box engage which is also equal to the beginning available energy (KE of the Box) minus the work done before engaging the spring. Is that clear?